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leonid [27]
3 years ago
15

Calculate the wavelength of a body of mass 96.0 mg and moving with a velocity of 30.0 m/s. The value of Planck’s constant is 6.6

26 × 10–34 J s.
Chemistry
1 answer:
Katen [24]3 years ago
6 0

Answer:

vcvvvvvvvvvvvvvv

Explanation:

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depth ; the distance from the top to the bottom of something .
geography ; the study of the earth s  physical features and the people ,plants , and animals.
light ; brightness from the sun or from a light.
medium ; between small and  large in   size.
period ;  the name given to a horizontal row of the periodic table.
properties ; attributes of a substance.
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intertidal ; relating to the region    between the hide tide mark and the low  tide mark .
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epipelagic ; relating to or inhabiting the uppermost layer of the water column of the open ocean  , into which enough sunlight enters for photosynthesis to take place.
mesopelagic ; relating to or inhabiting the layer of the water column in the open sea that lies between the epipelagic   and bathypelagic  layers at depths of about 200 to 1,000 meters.
bathypelagic ; relating to or inhabiting the layer of the water column  of the open sea that lies between the mesopelagic  and abyssopelagic  layers at depths of about 1,000 to 4,000.
abyssopelagic ; refering to or occurring in the region of deep water above the floor of the ocean .
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4 0
3 years ago
A container has a mixture of NO2 gas and N2O4 gas in equilibrium. The chemical reaction between the two gases is described by th
kondaur [170]

Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = \frac{38.8}{61.2^{2} }

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

0.0104 = \frac{200 - P(NO_{2})  }{[P(NO_{2} )]^{2}}

0.0104[P(NO_{2} )]^{2} + P(NO_{2} ) - 200 = 0

Resolving the second degree equation:

P(NO_{2} ) = \frac{-1+\sqrt{9.32} }{0.0208}

P(NO_{2} ) = 98.7

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are P(NO_{2} ) = 98.7 MPa and P(N₂O₄) = 101.3 MPa

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